Ultimately, we are headed towards a characterization of formal smoothness for reasonable morphisms (e.g. the types one encounters in classical algebraic geometry): we want to show that they are precisely the flat morphisms whose fibers are smooth varieties. This will be a much more usable criterion in practice (formal smoothness is given by a somewhat abstract lifting property, but checking that a concrete variety is smooth is much easier).  This is the intuition between smoothness: one should think of a flat map is a “continuously varying” family of fibers, and one wishes the fibers to be regular. This corresponds to the fact from differential topology that a submersion has submanifolds as its fibers.

It is actually far from obvious that a formally smooth (and finitely presented) morphism is even flat. Ultimately, the idea of the proof is going to be write the ring as a quotient of a localization of a polynomial ring. The advantage is that this auxiliary ring will be clearly flat, and it will also have fibers that are regular local rings.  In a regular local ring, we have a large supply of regular sequences, and the point is that we will be able to lift the regularity of these sequences from the fiber to the full ring.

Thus we shall use the following piece of local algebra.

Theorem Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a local homomorphism of local noetherian rings. Let ${M}$ be a finitely generated ${B}$-module, which is flat over ${A}$.

Let ${f \in B}$. Then the following are equivalent:

1. ${M/fM}$ is flat over ${A}$ and ${f: M \rightarrow M}$ is injective.
2. ${f: M \otimes k \rightarrow M \otimes k}$ is injective where ${k = A/\mathfrak{m}}$.

This is a useful criterion of checking when an element is ${M}$-regular by checking on the fiber. That is, what really matters is that we can deduce the first statement from the second.

Proof: All ${\mathrm{Tor}}$ functors here will be over ${A}$. If ${M/fM}$ is ${A}$-flat and ${f: M \rightarrow M}$ is injective, then the sequence

$\displaystyle 0 \rightarrow M \stackrel{f}{\rightarrow} M \rightarrow M/fM \rightarrow 0$

leads to a long exact sequence

$\displaystyle \mathrm{Tor}_1(k, M/fM) \rightarrow M \otimes k \stackrel{f}{\rightarrow} M \otimes k \rightarrow (M/fM) \otimes k \rightarrow 0.$

But since ${M/fM}$ is flat, it follows that ${M \otimes k \stackrel{f}{\rightarrow} M \otimes k}$ is injective.

The other direction is more subtle. Suppose multiplication by ${f}$ is a monomorphism on ${M \otimes k}$. Now write the exact sequence

$\displaystyle 0 \rightarrow P \rightarrow M \stackrel{f}{\rightarrow} M \rightarrow Q \rightarrow 0$

where ${P, Q}$ are the kernel and cokernel. We can also consider the image ${I = fM \subset M}$, to split this into two exact sequences

$\displaystyle 0 \rightarrow P \rightarrow M \rightarrow I \rightarrow 0$

and

$\displaystyle 0 \rightarrow I \rightarrow M \rightarrow Q \rightarrow 0.$

Here the map ${M \otimes k \rightarrow I \otimes k \rightarrow M \otimes k}$ is given by multiplication by ${f}$, so it is injective by hypothesis. This implies that ${M \otimes k \rightarrow I \otimes k}$ is injective. So ${M \otimes k \rightarrow I \otimes k}$ is actually an isomorphism because it is obviously surjective, and we have just seen it is injective. Next, since ${M \rightarrow I}$ is surjective, ${I \otimes k \rightarrow M \otimes k}$ is also injective.

(Given a sequence ${X \rightarrow Y \rightarrow Z}$ such that ${X \rightarrow Y}$ is surjective, ${X \rightarrow Z}$ is injective, it follows that ${Y \rightarrow Z}$ is itself injective: for if a nonzero ${y \in Y}$ mapped to zero in ${Z}$, then ${y}$ is hit by something in ${X}$, nonzero. Then that would go to zero in ${Z}$, contradiction.)

Let us tensor these two exact sequences with ${k}$. We get

$\displaystyle 0 \rightarrow \mathrm{Tor}_1(k, I) \rightarrow P \otimes k \rightarrow M \otimes k \rightarrow I \otimes k \rightarrow 0$

because ${M}$ is flat. We also get

$\displaystyle 0 \rightarrow \mathrm{Tor}_1(k, Q) \rightarrow I \otimes k \rightarrow M \otimes k \rightarrow Q \otimes k \rightarrow 0 .$

We’ll start by using the second sequence. Now ${I \otimes k \rightarrow M \otimes k}$ was just said to be injective, so that ${\mathrm{Tor}_1(k, Q) = 0}$. By the local criterion for flatness (Theorem 4.2 in the chapter on flatness in the CRing project), it follows that ${Q}$ is a flat ${A}$-module as well. But ${Q = M/fM}$, so this gives one part of what we wanted.

Now, we want to show finally that ${P = 0}$. Now, ${I}$ is flat; indeed, it is the kernel of a surjection of flat maps ${M \rightarrow Q}$, so the long exact sequence shows that it is flat. So we have a short exact sequence

$\displaystyle 0 \rightarrow P \otimes k \rightarrow M \otimes k \rightarrow I \otimes k \rightarrow 0,$

which shows now that ${P \otimes k = 0}$ (as ${M \otimes k \rightarrow I \otimes k}$ was just shown to be injective earlier). By Nakayama ${P = 0}$. This implies that ${f}$ is ${M}$-regular.

Corollary 7 Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a morphism of noetherian local rings. Suppose ${M}$ is a finitely generated ${B}$-module, which is flat over ${A}$.Let ${f_1, \dots, f_k \in \mathfrak{n}}$. Suppose that ${f_1, \dots, f_k}$ is a regular sequence on ${M \otimes k}$. Then it is a regular sequence on ${M}$ and, in fact, ${M/(f_1, \dots, f_k ) M}$ is flat over ${A}$.

Proof: This is now clear by induction.